TBR Physics Question - #4 of Review Passage I in Chapter 1

[Chemking]

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I have a question here that is really racking my brain. I am not completely following how the solution was written to this question. Thanks in advance:

A cannon is located on the edge of a cliff 100m above a flat, sandy canyon floor. three cannon balls are launched from the cannon with the same initial speed of 50 m/s, but at different angles. the cannon balls are identical except for their color. A red ball is launched horizontally, a white ball is launched at a 30 degree angle above the horizontal, and a blue ball is launched at a 60 degree angle above the horizontal.

You may ignore air resistance and the length of the cannon, unless other wise stated.

For reference: sin of 30 degrees = .5 Cos 30 degrees = .86
Sin 60 degrees = .86 Cos 60 degrees = .50
g = 10 m/s^2


Approximately how far above the canyon floor will the white ball be when it reaches its apex?
a) 30m
b) 120 m
c) 131 m
d) 225 m

 

[shefv]

What is supposed to be the correct answer? Is it C?

I can post my solution but want to check that I did it right first.

 

[Chemking]

Yep, C is correct

 

[Cawolf]

First find the vertical component of the velocity for the white ball.

vz = (50 m/s)(sin 30) = (50 m/s)(0.5) = 25 m/s

Then rearrange the kinematics equation to find when the velocity upwards is zero, as this is when the ball reaches the apex, which is what you want to find.

Without typing out the whole derivation (if you aren't comfortable with formula manipulations then this is worth memorizing).

h = v^2/2g = (25 m/s)^2/(2*10 m/s^2) = (625 m^2/s^2)(20 m/s^2) = (625/20) m = 31.25 m

This gives us the height of the ball above the launch point, which is 100 m above the canyon floor, so we add them.

h = 100 m + 31.25 m = 131.25 m

 

[Labrat07]

The equation @Cawolf used is v^2=v^2 + 2ah.

For the calculation, I'd estimate >30 but close then add 100.